Question

como acho o resultado dessa conta usando bhaskarax²-11y=45 x=2y

Answer 1

$x=2y$
$y= \frac{x}{2}$

$x^2-11y=45$
$x^2-11* \frac{x}{2} =45$
$x^2- \frac{11}{2} x-45=0$
$(a=1;b= -\frac{11}{2} ;c=-45)$

$\Delta = b^2-4ac$
$\Delta = (- \frac{11}{2}) ^2-4*1*(-45)$
$\Delta = \frac{121}{4} +180$
$\Delta = \frac{121+4*(180)}{4}$
$\Delta = \frac{121+720}{4}$
$\Delta = \frac{841}{4}$

$x= \frac{-b \pm \sqrt{\Delta} }{2a}$
$x= \frac{-(- \frac{11}{2}) \pm \sqrt{ \frac{841}{4} } }{2*1}$
$x= \frac{\frac{11}{2} \pm \sqrt{ \frac{29^2}{2^2} } }{2}$
$x= \frac{\frac{11}{2} \pm \frac{29}{2} }{2}$
$x= \frac{\frac{11\pm 29}{2}}{2}$
$x=\frac{11\pm 29}{4}}$

$\left \{ {{x_1= \frac{11+29}{4}= \frac{40}{4}=10 } \atop {x_2= \frac{11-29}{4}= \frac{-18}{4}= \frac{-9}{2}=-4.5 }} \right.$

$\left \{ {{y_1= \frac{x_1}{2}= \frac{10}{2}=5 } \atop {y_2= \frac{x_2}{2}= \frac{-4.5}{2}=-2.25 }} \right.$

$S=\{(10;5);(-4.5;-2.25)\}$